A set of points $\{P_0,P_1,P_2\}$ in $\mathbb{R}^2$ is called geometrically independent if the set of vectors $\{P_1-P_0, P_2-P_0\}$ is linearly independent.
Given a point $P=(x_1,y_1)$ and $r>0$ the boundary of a square with center $P$ and "radius" $r$ which has sides aligned to $xy$-axes is formally defined as the set $\{(x,y)\in\mathbb{R}^2 \ | \ max(|x-x_1|,|y-y_1|)=r\}$.
Given distinct points $P_0, P_1, P_2$ in $\mathbb{R}^2$ such that $\{P_0, P_1, P_2\}$ are geometrically independent it is a well known fact that there exists a unique circle $\mathcal{C}$ passing through points $P_0,P_1,P_2$.
Does this claim also hold if we consider $\mathcal{C}$ to be a boundary of a square which has its sides aligned to the $xy$-axes?
No.
$(0,0)$, $(1,0)$ and $(0,1)$ are geom. independent, but there exist infinitely many such squares passing through them.